5 ms·

> Since from an observers POV nothing can actually fall into the black hole, no black hole can form. The event horizon isn't really a physical boundary in that

> Since from an observers POV nothing can actually fall into the black hole, no black hole can form. The event horizon isn't really a physical boundary in that sense. It's the mathematical boundary at which, according to general relativity, a particle must have velocity equal to the speed of light in order to escape. A density change inside the star can change the size and shape of that boundary without things falling into it in the usual sense. In the same way, a density wave can 'travel' faster than the speed of sound (or light) in a medium because the wave is a mathematical construct that emerges from a physical situation.

grkvlt 9y ago
A better way of thinking about the event horizon is that it is the boundary beyond which events happening cannot be assigned a 'when' in our universe - i.e. the object falling 'in' to the black hole is never seen to pass this horizon, but to the object it seems that they do, however those events happen to it after an infinite time in our universe, i.e. never, due to time dilation. The contents of a black hole are the events that occur outside/after our universe relative to that boundary. I found these explanations on an awesome PBS youtube video [0] about black holes and space-time. 0. https://youtu.be/vNaEBbFbvcY https://youtu.be/vNaEBbFbvcY
criddell 9y ago
> a particle must have velocity equal to the speed of light in order to escape This has always confused me. Does a photon have mass? I've always thought the answer is no and so I don't understand why even light can't escape from a black hole.
- Santosh83 9y ago
  It's a consequence of the bending of spacetime around a singularity. All possible paths that light (or anything else) can take through spacetime lead towards the singularity.
  - criddell 9y ago
    So it's the same mechanism as gravitational lensing (or whatever it's called) that bends the path a photon takes when it passes something massive?
    - Santosh83 9y ago
      Yes, except here its much more extreme so that inside a certain distance (event horizon) spacetime is bent in such a way that all paths through it lead towards the singularity. That's the theory at least.
- deleted 9y ago
  [deleted]
- raattgift 9y ago
  In spacetime there are paths of least resistance called geodesics, and an object left alone will bind to a geodesic determined by the distribution of moving masses in the spacetime. If we take two parallel geodesics in empty spacetime and draw (a section of each of) them like this ||. But let's consider if we put a massive object like a star (O) somewhere near the geodesics. We'll exaggerate in the diagrams: O<| vs |>O vs >O< vs | O | etc, depending on where we put the star in relation to the two geodesics. Note that if they are close enough, they bend towards the star. Now we just have to bind an object to one of these ten geodesics shown schematically above. The strong equivalence principle stems from the observations by Galileo et al. that objects of different weights and configurations fall at the same rate (if one can eliminate air drag and so on). Any object may bind to an available geodesic, whether it's a feather, a bowling-ball, a beam of light, or a moon. One has to do work to move an object off a geodesic [1]. That light binds to geodesics and geodesics are determined by proximity to mass was tested by Eddington et al. during the 1919 solar eclipse, where they observed something similar to the |>O diagram above. Gravitational lensing works the same way. As we increase the mass of O, the closer geodesics are more and more bent towards O. So for a lighter star: |)o Black holes are much more massive (and yet more compact) than O, so there are geodesics more bent towards the black hole (because of the mass) and and more geodesics closer to the black hole's centre of mass. The closer geodesics can be bent around the black hole, possibly several times. Additionally there are "no return" geodesics that twist into circular orbits around the black hole. There is an innermost stable circular orbit (ISCO) too. Finally, there are "no return" geodesics that lead past the ISCO and into the region covered by the event horizon. @ | could be a diagram where we replace O in )O | with a black hole. Light can bind to any of these "no return" geodesics just like any other object like a feather or a bowling ball. - -- [1] Strictly speaking, our universe is 1+3 Lorentzian with extremely high experimental confidence. One dimension is timelike and the other three spacelike. This lets us sort geodesics into three types: spacelike, timelike, and null (or lightlike). In normal empty space light (and any other massless particle) always moves along a null geodesic, and moving it off a null geodesic is energetically impossible. Likewise, in normal empty space, massive particles always move along timelike geodesics, and while (with a lot of work) you can move them onto timelike geodesics that look more and more lightlike, it's energetically impossible to push it onto a lightlike geodesic. Distinguishing between lightlike and timelike is best done with respect to some coordinates, intervals, and using a tiny bit of calculus. The Euclidean distance for an object only moving in one spatial direction is ds^2 = dx^2. The spacetime interval for an object only moving in the timelike direction is ds^2 = c^2dt^2. If we let it move in the x direction, it's ds^2 = c^2dt^2 - dx^2. For light, and units of lightseconds in x and seconds in t, we have ds^2 = 0, thus "null". If ds^2 > 0, the interval is timelike. If between every two points on a geodesic the interval is lightlike, the geodesic is lightlike. If between every two points on a geodesic the interval is timelike, the geodesic is timelike: an object bound to such a geodesic does not travel as far in space over a given time as light does. The most lightlike but still timelike geodesic is available to ultra-relativistic massive objects. So if we define an event horizon as the surface below which all lightlike geodesics lead inward, we have also forced ultra-relativistic massive objects inwards on their almost-lightlike geodesics. Putting this more colloquially, if you are inside the event horizon, even if you could accelerate to the speed of light, you aren't getting out.
- pdonis 9y ago
  > Does a photon have mass? It has zero rest mass, but a photon is never at rest. Photons have energy, and anything that has energy is subject to gravity in general relativity.

grkvlt 9y ago

A better way of thinking about the event horizon is that it is the boundary beyond which events happening cannot be assigned a 'when' in our universe - i.e. the object falling 'in' to the black hole is never seen to pass this horizon, but to the object it seems that they do, however those events happen to it after an infinite time in our universe, i.e. never, due to time dilation. The contents of a black hole are the events that occur outside/after our universe relative to that boundary. I found these explanations on an awesome PBS youtube video [0] about black holes and space-time. 0. https://youtu.be/vNaEBbFbvcY https://youtu.be/vNaEBbFbvcY

criddell 9y ago

> a particle must have velocity equal to the speed of light in order to escape This has always confused me. Does a photon have mass? I've always thought the answer is no and so I don't understand why even light can't escape from a black hole.

Santosh83 9y ago
It's a consequence of the bending of spacetime around a singularity. All possible paths that light (or anything else) can take through spacetime lead towards the singularity.
- criddell 9y ago
  So it's the same mechanism as gravitational lensing (or whatever it's called) that bends the path a photon takes when it passes something massive?
  - Santosh83 9y ago
    Yes, except here its much more extreme so that inside a certain distance (event horizon) spacetime is bent in such a way that all paths through it lead towards the singularity. That's the theory at least.
deleted 9y ago
[deleted]
raattgift 9y ago
In spacetime there are paths of least resistance called geodesics, and an object left alone will bind to a geodesic determined by the distribution of moving masses in the spacetime. If we take two parallel geodesics in empty spacetime and draw (a section of each of) them like this ||. But let's consider if we put a massive object like a star (O) somewhere near the geodesics. We'll exaggerate in the diagrams: O<| vs |>O vs >O< vs | O | etc, depending on where we put the star in relation to the two geodesics. Note that if they are close enough, they bend towards the star. Now we just have to bind an object to one of these ten geodesics shown schematically above. The strong equivalence principle stems from the observations by Galileo et al. that objects of different weights and configurations fall at the same rate (if one can eliminate air drag and so on). Any object may bind to an available geodesic, whether it's a feather, a bowling-ball, a beam of light, or a moon. One has to do work to move an object off a geodesic [1]. That light binds to geodesics and geodesics are determined by proximity to mass was tested by Eddington et al. during the 1919 solar eclipse, where they observed something similar to the |>O diagram above. Gravitational lensing works the same way. As we increase the mass of O, the closer geodesics are more and more bent towards O. So for a lighter star: |)o Black holes are much more massive (and yet more compact) than O, so there are geodesics more bent towards the black hole (because of the mass) and and more geodesics closer to the black hole's centre of mass. The closer geodesics can be bent around the black hole, possibly several times. Additionally there are "no return" geodesics that twist into circular orbits around the black hole. There is an innermost stable circular orbit (ISCO) too. Finally, there are "no return" geodesics that lead past the ISCO and into the region covered by the event horizon. @ | could be a diagram where we replace O in )O | with a black hole. Light can bind to any of these "no return" geodesics just like any other object like a feather or a bowling ball. - -- [1] Strictly speaking, our universe is 1+3 Lorentzian with extremely high experimental confidence. One dimension is timelike and the other three spacelike. This lets us sort geodesics into three types: spacelike, timelike, and null (or lightlike). In normal empty space light (and any other massless particle) always moves along a null geodesic, and moving it off a null geodesic is energetically impossible. Likewise, in normal empty space, massive particles always move along timelike geodesics, and while (with a lot of work) you can move them onto timelike geodesics that look more and more lightlike, it's energetically impossible to push it onto a lightlike geodesic. Distinguishing between lightlike and timelike is best done with respect to some coordinates, intervals, and using a tiny bit of calculus. The Euclidean distance for an object only moving in one spatial direction is ds^2 = dx^2. The spacetime interval for an object only moving in the timelike direction is ds^2 = c^2dt^2. If we let it move in the x direction, it's ds^2 = c^2dt^2 - dx^2. For light, and units of lightseconds in x and seconds in t, we have ds^2 = 0, thus "null". If ds^2 > 0, the interval is timelike. If between every two points on a geodesic the interval is lightlike, the geodesic is lightlike. If between every two points on a geodesic the interval is timelike, the geodesic is timelike: an object bound to such a geodesic does not travel as far in space over a given time as light does. The most lightlike but still timelike geodesic is available to ultra-relativistic massive objects. So if we define an event horizon as the surface below which all lightlike geodesics lead inward, we have also forced ultra-relativistic massive objects inwards on their almost-lightlike geodesics. Putting this more colloquially, if you are inside the event horizon, even if you could accelerate to the speed of light, you aren't getting out.
pdonis 9y ago
> Does a photon have mass? It has zero rest mass, but a photon is never at rest. Photons have energy, and anything that has energy is subject to gravity in general relativity.